If ${\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3,\dots ,{\mathrm{a}}_{24}$ are  in arithmetic progression  and  ${\mathrm{a}}_1+{\mathrm{a}}_5+{\mathrm{a}}_{10}+{\mathrm{a}}_{15}+{\mathrm{a}}_{20}+{\mathrm{a}}_{24}=225$ then ${\mathrm{a}}_1+{\mathrm{a}}_2+{\mathrm{a}}_3+\dots +{\mathrm{a}}_{23}+{\mathrm{a}}_{24}$ is equal to

Solution:

Given that,

  ${\mathrm{a}}_1+{\mathrm{a}}_5+{\mathrm{a}}_{10}+{\mathrm{a}}_{15}+{\mathrm{a}}_{20}+{\mathrm{a}}_{24}=225$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left({\mathrm{a}}_1+{\mathrm{a}}_{24}\right)+\left({\mathrm{a}}_5+{\mathrm{a}}_{20}\right)+\left({\mathrm{a}}_{10}+{\mathrm{a}}_{15}\right)=225\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\left({\mathrm{a}}_1+{\mathrm{a}}_{24}\right)=225\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{a}}_1+{\mathrm{a}}_{24}=75\end{array}$

[$\because$in an AP the sum of the terms equidistant from the beginning and the end is same and is equal to the sum of the first and last term]

 $\begin{array}{l}\therefore {\mathrm{a}}_1+{\mathrm{a}}_2+\dots +{\mathrm{a}}_{24}=\frac{\mathrm{n}}{2}(\mathrm{a}+\mathrm{l})=\frac{24}{2}\left({\mathrm{a}}_1+{\mathrm{a}}_{24}\right)\\ =12\times 75=900\end{array}$ [from eq.(i)]